ĐKXĐ:\(x\ge-6\)
Ta có:
\(x^2-2x+2\\
=\left(x^2-2x+1\right)+1\\
=\left(x-1\right)^2+1>0\\
\Rightarrow\left|x^2-2x+2\right|=x^2-2x+2\)
\(\left|x^2-2x+2\right|=x+6\\ \Leftrightarrow x^2-2x+2-x-6=0\\ \Leftrightarrow x^2-3x-4=0\\ \Leftrightarrow x^2+x-4x-4=0\\ \Leftrightarrow x\left(x+1\right)-4\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\left(tm\right)\\x=4\left(tm\right)\end{matrix}\right.\)
Vậy pt có tập nghiệm `S={-1;4}`